1. **State the problem:** We need to find the distance between two point charges with magnitudes $+150$ nC and $+250$ nC given that the electric force between them is $20$ N. 2. **Formula used:** The electric force between two point charges is given by Coulomb's law: $$F = k \frac{|q_1 q_2|}{r^2}$$ where $F$ is the force, $k = 8.99 \times 10^9$ N·m²/C² is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. 3. **Convert charges to Coulombs:** $$q_1 = 150 \text{ nC} = 150 \times 10^{-9} = 1.5 \times 10^{-7} \text{ C}$$ $$q_2 = 250 \text{ nC} = 250 \times 10^{-9} = 2.5 \times 10^{-7} \text{ C}$$ 4. **Rearrange formula to solve for $r$:** $$r^2 = k \frac{|q_1 q_2|}{F}$$ 5. **Substitute values:** $$r^2 = 8.99 \times 10^9 \times \frac{(1.5 \times 10^{-7})(2.5 \times 10^{-7})}{20}$$ 6. **Calculate numerator:** $$(1.5 \times 10^{-7})(2.5 \times 10^{-7}) = 3.75 \times 10^{-14}$$ 7. **Calculate $r^2$:** $$r^2 = 8.99 \times 10^9 \times \frac{3.75 \times 10^{-14}}{20} = 8.99 \times 10^9 \times 1.875 \times 10^{-15}$$ 8. **Multiply:** $$r^2 = 1.685625 \times 10^{-5}$$ 9. **Take square root:** $$r = \sqrt{1.685625 \times 10^{-5}} = 0.004106 \text{ meters}$$ 10. **Convert meters to centimeters:** $$r = 0.004106 \times 100 = 0.4106 \text{ cm}$$ **Final answer:** The two charges are approximately $0.41$ centimeters apart.